larutan glukosa (mr=180 g mol-1) dalam air membeku pada suhu -0,372°c m-1 , jumlah glukosa yang terlarut dalam 500 g air adalah... a. 9 gr b. 36 gr c. 18 gr d.

Mr = 180 gram/mol
Tf = - 0,372°C
wp = 500 gram
Kf = 1,86°C
wt = ..............?

ΔTf = Tf° - Tf
       =  0 - (- 0,372)
       = 0,372°C

ΔTf = m.Kf
ΔTf = wp/Mr x 1000/wp x Kf
0,372 = wp/180 x 1000/500 x 1,86
0,372 = wp/180 x 3,72
wp = (180 x 0,372)/3,72
wp = 180 x 0,1
wp = 18 gram

"jawaban : C"